题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3555

Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)

Total Submission(s): 15270    Accepted Submission(s): 5515

Problem Description

The counter-terrorists found a time bomb in the dust. But this time the terrorists improve on the time bomb. The number sequence of the time bomb counts from 1 to N. If the current number sequence includes the sub-sequence "49", the power of the blast would add one point. Now the counter-terrorist knows the number N. They want to know the final points of the power. Can you help them?  

Input

The first line of input consists of an integer T (1 <= T <= 10000), indicating the number of test cases. For each test case, there will be an integer N (1 <= N <= 2^63-1) as the description. The input terminates by end of file marker.  

Output

For each test case, output an integer indicating the final points of the power.  

Sample Input

3 1 50 500  

Sample Output

0 1 15 Hint From 1 to 500, the numbers that include the sub-sequence "49" are "49","149","249","349","449","490","491","492","493","494","495","496","497","498","499", so the answer is 15.  

Author

fatboy_cw@WHU  

Source

2010 ACM-ICPC Multi-University Training Contest（12）——Host by WHU  

Recommend

zhouzeyong

思路：经典的数位DP。自己不太懂这个知识点，参考了一位大神的博客，很详细，如下所示。

http://www.cnblogs.com/liuxueyang/archive/2013/04/14/3020032.html

附上AC代码：

#include <bits/stdc++.h> using namespace std; typedef long long ll; const int maxn = 25; ll dp[maxn][3]; int num[maxn]; ll n; void init(){ dp[0][0] = 1; for (int i=1; i<maxn; ++i){ dp[i][0] = dp[i-1][0]*10-dp[i-1][1]; dp[i][1] = dp[i-1][0]; dp[i][2] = dp[i-1][2]*10+dp[i-1][1]; } } int main(){ init(); int T; scanf("%d", &T); while (T--){ scanf("%I64d", &n); ll ans = 0; bool ok = false; int len = 0; while (n){ num[++len] = n%10; n /= 10; } for (int i=len; i>=1; --i){ ans += dp[i-1][2]*num[i]; if (ok) ans += dp[i-1][0]*num[i]; if (!ok && num[i]>4) ans += dp[i-1][1]; if (i+1<=len && num[i+1]==4 && num[i]==9) ok = true; } if (ok) ++ans; printf("%I64d\n", ans); } return 0; }